How to solve this length integral?

Question

I have to solve this integral in order to calculate the length of a curve

$L(\gamma) = 2\alpha\int_0^L \sqrt{e^t +1} dt$

I tried with some substitutions, like $e^t +1 = u$ and others, but the calculations get difficult and the result I get is different from the book's one ($L(\gamma) = 2\sqrt{2}\alpha sinh(L)$)

Answer 1

Let $u = \sqrt{e^t + 1}$ following the comment by @Gribouillis

$t=\log \left(u^2-1\right)$

$dt = \dfrac{2 u \,du}{u^2-1}$

The integral becomes

$$L(\gamma)=4 a \int_{\sqrt{2}}^{\sqrt{e^L+1}} \dfrac{u^2}{u^2-1} \, du$$

$$\int \frac{u^2}{u^2-1} \, du=\int \frac{\left(u^2-1\right)+1}{u^2-1} \, du=\int \left(\frac{1}{u^2-1}+1\right) \, du=$$

$$=u+\frac{1}{2} \log \left(\frac{u-1}{u+1}\right)$$

Therefore $$L(\gamma)=4 a \left[u+\frac{1}{2} \log \left(\frac{u-1}{u+1}\right)\right]_{\sqrt{2}}^{\sqrt{e^L+1}}=4 a \left(\sqrt{e^L+1}+\frac{1}{2} \log \left(\frac{\sqrt{e^L+1}-1}{\sqrt{e^L+1}+1}\right)-\sqrt{2}-\frac{1}{2} \log \left(\frac{\sqrt{2}-1}{1+\sqrt{2}}\right)\right)=2 a \left(2 \sqrt{e^L+1}+\log \left(\sqrt{e^L+1}-1\right)-\log \left(\sqrt{e^L+1}+1\right)\\-2 \sqrt{2}-\log \left(\sqrt{2}-1\right)+\log \left(1+\sqrt{2}\right)\right)$$

Answer 2

I can simplify the solution of @Raffaele in the sense of reducing the clutter. So let's take it from

$$ \begin{align} \int \frac{u^2}{u^2-1} &=u+\frac{1}{2} \log \left(\frac{u-1}{u+1}\right)\\ &=u-\tanh^{-1}u \end{align} $$

Then the complete solution is given by

$$ \begin{align} L(\gamma) &= 2\alpha\int_0^L \sqrt{e^t +1} dt\\ &=4\alpha\left[\sqrt{e^L + 1}-\sqrt{2}-\tanh^{-1}\left(\sqrt{e^L + 1}\right) +\tanh^{-1}\sqrt{2} \right]\\ \end{align} $$

I have verified this solution by comparison with direct numerical simulation of the integral. In addition, I can say unequivocally that the answer profferred from the book is erroneous.